# Proving that $\sqrt 3\not\in Q(\sqrt[4]2)$ [duplicate]

I came across this problem while solving another one. I will show how far I could get on my own:

Suppose that $$\sqrt 3 \in Q(\sqrt[4]2)$$. Since $$Q(\sqrt3)$$ is the smallest field containing both $$Q$$ and $$\sqrt3$$, thus $$Q(\sqrt3)\subset Q(\sqrt[4]2)$$.

Now $$[Q(\sqrt[4]2):Q] =4$$ and $$[Q(\sqrt3):Q]=2$$ (I've already proven both), so$$[Q(\sqrt[4]2):Q(\sqrt 3)] =2$$. Therefore by the definition of degree extension, there exist $$p(x)\in Q(\sqrt 3)[X]$$ of degree 2 realizing $$\sqrt[4]2$$, that is, exists: $$a,b,c\in Q(\sqrt 3)$$ such that:

$$a(\sqrt[4]2)^2+b \sqrt[4]2+c =0.$$

How to proceed now? I should use the fact that those extensions lie in $$\mathbb R$$ to get a contradiction? I only know that $$a\neq 0$$, but what about the other coefficients?

EDIT: I could find an argument below, check the answers. Check it out if you agree! The argument is just a continuation of the above one.

## marked as duplicate by Watson, Community♦Nov 23 '18 at 21:22

• Possibly related: math.stackexchange.com/questions/2995190. A computation-free proof would be : if $\sqrt 3 \in \Bbb Q(\sqrt[4]{p})$ for some prime $p \neq 3$, then we have an inclusion $\Bbb Q(\sqrt p, \sqrt 3) \subset \Bbb Q(\sqrt[4]{p})$, which is an equality because of the degrees (one only needs a small computation with the traces to show that the former field has indeed degree $4$ over $\Bbb Q$ ; the latter has degree $4$ by Eisenstein). […] – Watson Nov 23 '18 at 15:56
• […] Then $\Bbb Q(\sqrt[4]{p})$ would be Galois over $\Bbb Q$, since $\Bbb Q(\sqrt p, \sqrt 3) / \Bbb Q$ is. But the conjugate $i \sqrt[4]{p}$ of $\sqrt[4]{p}$ does not belong to $\Bbb Q(\sqrt[4]{p}) \subset \Bbb R$. – Watson Nov 23 '18 at 15:56
• (My proof should also show that $\sqrt n \not \in \Bbb Q(\sqrt[4]{m})$ for every coprime integers $n,m>1$). – Watson Nov 23 '18 at 16:06
• @Watson That helped a lot! I used the method in that question to solve it. – math.h Nov 23 '18 at 21:22
• I asked a similar question before math.stackexchange.com/questions/2431221/…. This might be helpful. – Seewoo Lee Nov 24 '18 at 3:23

I think that I could find an answer. Check it out if it's correct:

As $$p(x) = ax^2+bx+c\in Q(\sqrt3)[X]$$ is a polynomial of smallest degree killing $$\sqrt[4]2$$ (by the very definition of being algebraic of degree 2 over $$Q(\sqrt3)$$), we must have $$a\neq 0$$. Also we can conclude that $$c\neq 0$$, otherwise $$p(x) = xq(x)$$, where degree of $$q(x)$$ $$<$$ degree of $$p(x)$$; and therefore $$q(\sqrt[4]2) = 0,$$ which again contradicts the minimality of the degree of $$p(x)$$.

Now we see that $$b\neq 0$$. If not, noticing that those field extensions all lie in $$\mathbb R$$, we find that $$a\sqrt 2+c=0$$ and hence that $$\sqrt2\in Q(\sqrt3),$$ which does not happen.

Therefore, $$p(x)$$ is such that $$a,b,c$$ are all non-zero. Solving for $$\sqrt 2$$, we find that $$\sqrt2 = \frac{-b\sqrt[4]2-c}{a} \,.$$

Since $$\sqrt 2 \not\in Q(\sqrt 3)$$, we must have the denominator of the above equation equal to zero and hence $$\sqrt[4]2\in Q(\sqrt 3)$$. By the minimality argument of fields, we see that $$Q(\sqrt 3) = Q(\sqrt[4]2)$$ (noticed that we have already assumed that $$Q(\sqrt 3) \subset Q(\sqrt[4]2)$$). But again this cannot happen, since the degree of $$Q(\sqrt[4]2)$$ over $$Q$$ is $$4$$ and the degree of $$Q(\sqrt 3)$$ over $$Q$$ is $$2$$.

With all this such contradictions, we conclude that $$\sqrt 3 \not \in Q(\sqrt[4]2)$$.

$$\sqrt3=p/q\sqrt[4]{2} \text{ with } gcd(p,q)=1\Rightarrow 9=(p^4/q^4) 2\Rightarrow 9q^4=2p^4\Rightarrow 3|p \text{ and } 2|q.$$ Suppose $$p=3a$$ and $$q=2b$$. Then $$9\times 16 b^4=2\times 81 a^4\Rightarrow 8b^4=9a^4\Rightarrow 3|b\Rightarrow 3|q\Rightarrow gcd(p,q)>1.$$